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A multi-variable problem

  1. Apr 4, 2008 #1
    My problem is this.
    We have three javelin throwers A,B and C.
    It is known that A defeats B with probability 60%, B defeats C with probability again 60% and A has the better of C with 70%.
    What are the a-priori probabilities of A, B, C winning the 3-way contest ?

    If in general we have N participants and Pij for the head to head probabilities, what is the formula for the win probability of each (Qi = some f(Pij)) ?
  2. jcsd
  3. Apr 5, 2008 #2


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    Welcome to PF!

    Hi plymouth! Welcome to PF! :smile:

    Show us what you've tried so far, and then we'll know how to help you! :smile:
  4. Apr 5, 2008 #3
    Well I 'm a little baffled.
    I suppose -in real life- those percentages 60%, 60%, 70% might derive from statistical observations and logbooks.
    So if A throws 85 meters plus-minus 2 metres and if B throws 84.5 meters plus-minus 2 metres -as measured by the sports statisticians- that gives us the percentage of A versus B, using the error integral, and let's say that comes out to be 60%.
    Similarly for the other two pairs B-C and A-C.
    That type of work takes us to a different sort of sample space -the Gauss distribution- but now the maths are becoming exceedingly awkward.

    You could say that A has probability 0.6 x 0.7 = 0.42, B has 0.4 x 0.6 = 0.24 and C has 0.3 x 0.4 = 0.12 but those values add up to 0.88 instead of 1.00 so it's only a crude approximation.

    There has to be a better approximation somewhere.

    One could also say that since B throws on the average 0.5 meters less than A -the record holder in the company- then the probability might be something of the order of c x exp ( - 0.5 / d ), where d is an experimental constant and c is a normalisation constant.
    This approach will, again, give you some not unrealistic figures for the probabilities in such events but that is only empirical.
    Last edited: Apr 5, 2008
  5. Apr 6, 2008 #4


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    Hi plymouth! :smile:

    Have you done Bayesian probability?
  6. Apr 6, 2008 #5
    I deleted a post because I wrote something in error.
    I have done Bayesian probability at college and some other things like the Poisson process.
    In this problem, in real life, we measure throw distances and therefore the normal distribution comes into play if you look at it from a rigorous point of view.
  7. Apr 6, 2008 #6


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    ok … unfortunately I know nothing about how to use Bayesian probability, so I can't comment on whether you should be using it. :redface:

    You'll have to wait for someone else to answer … or you could PM a homework mentor. :smile:
  8. Apr 6, 2008 #7
    Looks like I stated the problem from a different angle by talking in terms of Bayesian probabilities.
    In reality what we have is measurements of performences, then the normal distribution law comes into play, when we try to predict the winner. In this way only my approximate solution with the exp function works to a certain extent - it does n't look easy to better it, but I was wondering if some newer approximation exists.

    There are a multitude of sports situations corresponding to this problem not just the
    throwing of objects but also sprint races and even dog races.
    The failure of machines with many components may also be subject to the same law - I 'm not too sure about that.
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